Please help with conditional probability problem?
Over 50% of all engineers end up in management. An engineering firm needs project managers that understand the technical aspects of the product. A large firm is looking toward future needs. A look at personnel records indicates 30% of their current engineers (who are not yet in management positions) have an electrical engineering degree, 42% have a mechanical engineering degree,and 21% have a software engineering degree. 27% of all the engineers have some management training. Specifically, 11% have an electrical engineering degree and management training, 9% have a mechanical engineering degree and management training, and 2% have a software engineering degree and management training.
a)What is the probability that an engineer (any engineer) will not have had any management training?
b)What is the probability that an engineer with a mechanical engineering degree has had management training?
c)What is the probability an engineer will have some other degree or management training but not both
a) 1 - .27 = ?
However, the ones with some degree beside management training (27%) do not add up when the engineering degrees are summed (11% + 9% + 2% = 22%). What about the other 5%?
To solve this conditional probability problem, we will use the formula for conditional probability:
P(A | B) = P(A ∩ B) / P(B)
where P(A | B) represents the probability of event A occurring given that event B has already occurred, P(A ∩ B) represents the probability of both events A and B occurring, and P(B) represents the probability of event B occurring.
a) To find the probability that an engineer will not have had any management training (i.e., event A), we need to find the complement of the event "having management training" (i.e., event B).
P(A') = 1 - P(B)
The probability of having management training (P(B)) can be calculated by adding up the probabilities of engineers with each degree and management training:
P(B) = P(Electrical with management training) + P(Mechanical with management training) + P(Software with management training)
P(B) = 11% + 9% + 2% = 0.11 + 0.09 + 0.02 = 0.22
Therefore, P(A') = 1 - P(B) = 1 - 0.22 = 0.78
So, the probability that an engineer will not have had any management training is 0.78 or 78%.
b) To find the probability that an engineer with a mechanical engineering degree has had management training (i.e., event A given B), we can directly use the given probabilities:
P(A | B) = P(Mechanical with management training) / P(Mechanical)
First, we calculate P(Mechanical with management training) by using the given percentage:
P(Mechanical with management training) = 9% = 0.09
Next, we calculate P(Mechanical):
P(Mechanical) = P(Mechanical with management training) + P(Mechanical without management training)
The percentage of engineers with mechanical engineering degrees can be found from the given information. The engineers without management training can be calculated by subtracting the percentage of engineers with mechanical engineering degrees and management training from the percentage of engineers with mechanical engineering degrees:
P(Mechanical without management training) = P(Mechanical) - P(Mechanical with management training)
P(Mechanical) = 42% = 0.42
P(Mechanical without management training) = 0.42 - 0.09 = 0.33
Therefore, P(A | B) = P(Mechanical with management training) / P(Mechanical) = 0.09 / 0.42 ≈ 0.2143
So, the probability that an engineer with a mechanical engineering degree has had management training is approximately 0.2143 or 21.43%.
c) To find the probability that an engineer will have some other degree or management training but not both (event A ∪ B and not A ∩ B), we can use the following formula:
P(A ∪ B and not A ∩ B) = P(A ∪ B) - P(A ∩ B)
We already calculated P(A ∩ B) in part b. Now, we need to find P(A ∪ B) which represents the probability of an engineer having some other degree or management training.
P(A ∪ B) = P(Electrical) + P(Mechanical) + P(Software) - P(A ∩ B)
P(Electrical) = 30% = 0.30
P(Mechanical) = 42% = 0.42
P(Software) = 21% = 0.21
P(A ∩ B) = 0.09 (from part b)
P(A ∪ B) = 0.30 + 0.42 + 0.21 - 0.09 = 0.84
Therefore, P(A ∪ B and not A ∩ B) = P(A ∪ B) - P(A ∩ B) = 0.84 - 0.09 = 0.75
So, the probability that an engineer will have some other degree or management training but not both is 0.75 or 75%.
These are the solutions to the three conditional probability questions.